Question

A small sphere1 carrying charge q1 having inertia m1 is constrained to moving inside a narrow vertical. Fixed at the bottom of the tube is a small sphere 2 carrying charge q2.
Determine the quilibrium height h for sphere 1{ignore friction}

Express answers on terms of coulombs constant, acceleration due to gravity g and the variables q1. q2 and m1

Answer 1

There are two forces acting on the sphere 1: Coulomb's force acting up and gravitation acting down. The first one at height $h$ is given as follows:

$$F_C = k\dfrac{q_1q_2}{h^2}$$

The second one at any height (small compared with the size of the Earth) is given as follows:

$$F_g = m_1g$$

At equilibrium position they are equal to each other. Thus, obtain for height:

$$F_C = F_g\\$$ $$k\dfrac{q_1q_2}{h^2} = m_1g\\ kq_1q_2 = h^2m_1g\\ h = \sqrt{\dfrac{kq_1q_2}{m_1g}}$$

Answer. $h = \sqrt{\dfrac{kq_1q_2}{m_1g}}$.